Easy: Single Number

Problem Statement

In a non-empty array of integers, every number appears twice except for one, find that single number.

Example 1:

Input: 1, 4, 6, 1, 3, 6, 3
Output: 4

Example 2:

Input: 1, 9, 1
Output: 9

Solution

The simple solution is using a hashmap and iterate through the input:

if number already in the hashmap, remove it
if number is not present in hashmap, add it
in the end, only left 1 data in the hashmap

Time and space Complexity: Time Complexity of the above solution will be $O(n)$ and space complexity will also be $O(n)$ .

Can we do better than this using the XOR Pattern?

Recall the following two properties of XOR:

It returns zero if we take XOR of two same numbers.
It returns the same number if we XOR with zero. So we can XOR all the numbers in the input; duplicate numbers will zero out each other and we will be left with the single number.

Try it yourself https://go.dev/play/p/XcOwP-nIMDO

Code

package main
import (
	"testing"
)
func FindSingleNumber(list []int) int {
	num := 0
	for i := range list {
		num ^= list[i]
	}
	return num
}
func TestLastIndex(t *testing.T) {
	tests := []struct {
		list []int
		want int
	}{
		{list: []int{1, 4, 2, 1, 3, 2, 3}, want: 4},
		{list: []int{1, 9, 1}, want: 9},
	}
	for _, tt := range tests {
		if got := FindSingleNumber(tt.list); got != tt.want {
			t.Errorf("LastIndex(%v) = %v, want %v", tt.list, got, tt.want)
		}
	}
}

Time Complexity: Time complexity of this solution is $O(n)$ as we iterate through all numbers of the input once.

Space Complexity: The algorithm runs in constant space $O(1)$ .